Answer:
y maximum 3.54 m, value X 2.35 m
Explanation:
We have a projectile launch problem, let's calculate the maximum height of the projectile, where the vertical speed must be zero
Vyf² = Vyo² - 2g (Y-Yo)
Where Yo is the initial height of the ramp 1.5 m
0 = Vyo² -2g (Y-Yo)
Y-Yo = Voy² / 2g
Y = Yo + Voy² / 2g
Let's calculate the velocity components using trigonometry
Voy = vo without T
Vox = Vo cost
Voy = 7.3 sin 60
Vox = 7.3 cos 60
Voy = 6.32 m / s
Vox = 3.65 m / s
Let's calculate the maximum height
Y = 1.5 +6.32²/2 9.8
Y = 3.54 m
This is the maximum height from the ground
b) They ask us for the position of this point horizontally, we can calculate it looking for the time it took for the skateboarder to reach the highest point
Vfy = Voy - gt
0 = Voy - gt
t = Voy / g
t = 6.32 / 9.8
t = 0.645 s
Since there is no acceleration on the x-axis, we have a uniform movement, we can calculate the distance for this time
X = Vox t
X = 3.65 0.645
X= 2.35 m
Answer:
Magnitude = 4.056 m
Direction = 42.3⁰
Explanation:
The vector is resolved in terms of the vertical and horizontal components. Let's look each of these separately.
The vector 4.40 is directed East. This automatically becomes a horizontal component.
But we know that there is a vector 3.40 North West. The angle the vector makes with the horizontal is 61⁰.
Resolving the vectors should yield the horizontal and vertical components:
Horizontal components
The first component is 4.40 m
The second one is derived by resolving 3.40 to the horizontal like this 3.40 × - cos 61⁰ = -1.648 m
Adding the horizontal component gives 4.40 m + ( -1.648 m) = 2.752 m
Vertical components
Resolve 3.40 with the angle 61⁰ like this: vertical comp = 3.41 × sin 61
= 2.98 m
The magnitude is given by √[(2.98)²+ (2.752)²] = 4.056 m Ans
The direction us given by tan⁻¹ (2.98/2.752) = 42.3⁰ Ans
Answer: 0
Explanation: Initial velocity is 0.
Well idk if this helps but the formula to solve acceleration is
a=F/m=(100kg)=1.0m/s 2
